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• CommentRowNumber1.
• CommentTimeOct 16th 2021
• (edited Oct 16th 2021)

Let $p:X\rightarrow Y$ be a covering and $f:X \rightarrow Z$ be a map. What is the correct definition of the space of factorizations of $f$ through $Y$ up to homotopy ? I’d like a definition not as a space of maps $Y \rightarrow Z$ that makes the diagram homotopy commutative, but rather something like a spaces of homotopies from $f$ to a map that factors on the nose.. It’s probably super elementary but for some reasons it gets me confused.

Basically this is the topological picture generalizing: given an inclusion of groups $H \subset G$ and a morphism $H \rightarrow K$, classify the lifts $G\rightarrow K$ along this map (either on the nose, or up to conjugation, depending of whether we’re taking (un)pointed spaces and maps).

Now suppose that the normal closure of $H$ in $G$ is the whole of $G$. Say, to simplify, there is another subgroup $H'$ of $G$ which is conjugated to $H$ and such that $H,H'$ generate $G$. Now assume I also have a morphism $H' \rightarrow K$ (modulo details this is typically the sort of things I’d get from the topological picture, like if I have a morphism into $K$ out of the fundamental groupoid of $X$ with respect to the fiber over a chosen basepoint in $Y$, e.g. if i know alll those points are sent to the same point in $Z$). Now the question of whether these morphisms extends to a morphism from $G$ becomes a condition, not extra structure (ie we have a surjective map from the free product $H*H'$ to $G$ and a map from $H*H'$ to $K$ so now the question is whether it factors through a quotient (rather than extends along an inclusion)). In particular if it does factors through it does so in a unique way.

What is the correct way to model this topologically, or groupoidally ? In what setting can I use that the normal closure of $H$ is $G$ to get some sort of unicity of factorization of some map ? I guess what I’m really asking is whether thee is a name for a functor $F:C\rightarrow D$ such that $End(d)$ is generated by the images of the $End(c)$ for all $c$ s.t. $F(c)=d$ or something like that…

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeOct 16th 2021

Focusing on just the first paragraph (as I’d need to think more on what you say after “to simplify” :-)

The situation is this extension problem:

$\array{ X & \xrightarrow{f} & Z \\ {}^{\mathllap{cov}} \big\downarrow & \nearrow_{ \mathrlap{\exists ?} } \\ Y }$

I take it that the left map being a “covering” map is meant to say that it’s a principal bundle for some discrete group $S$? (Let me write $S$ for “structure group”.)

In that case, the vertical map is the quotient by the action of $S$. In the 1-category of topological spaces this means that the extension exists and then uiquely, if and only if $f$ is equivariant, i.e. here: invariant, since the corresponding $S$ action on $Z$ is taken to be trivial.

In the $\infty$-category of spaces the analogous statement holds in the homotopy coherent sense.

What you are after, I gather, is a hands-on 1-category-theoretic model for this homotopy coherent situation.

The model category for this purpose is the Borel model structure (which works with spaces modeled as simplicial sets).

Using the Quillen equivalence here says that the derived mapping space space of homotopically $S$-equivariant maps $f$ is equivalently the derived mapping space of the slice $sSet_{/\overline{W}S}$ from $Y \to \overline{W}S$ to $Z \times \overline{W} \to \overline{W}$ (here $\overline{W}S$ is the simplicial classifying space and $Y \to \overline{W}S$ is the map that classifies the covering), and this is equivalently the derived mapping space in $sSet$ from $Y$ to $Z$.

Now the $S$-action on $X$ is free, by assumption of it being a covering space, which by this Prop means that $X$ is cofibrant in the Borel model structure. Therefore the derived equivariant mapping space out of $X$ is modeled by the mapping complex of naively $S$-invariant maps.

In conclusion, up to homotopy the maps $f$ that do extend to $Y$ are those which have a naively $S$-invariant representative, and these then extend uniquely.

I think.

• CommentRowNumber3.
• CommentAuthorGuest
• CommentTimeOct 16th 2021

Hi Urs, thanks a lot for your comment. The thing is it’s precisely not a principal bundle for some discrete group, because in my case it’s not a regular covering. In other words, $\pi_1(X)$ is not normal in $\pi_1(Y)$ (and indeed the particular situation I have a hard time NPOV-ing is the situation where its normal closure is the whole of $\pi_1(Y)$). That being said, even in the situation you describe there’s a small subtlety (I’m cutting hairs but it got me confused for a while). Namely it’s not quite true that $f$ factors uniquely, in the sense that I can have two maps $s,t:Y\rightarrow Z$ which are not homotopic, but such that their precomposition with $p$ are both homotopic to $f$. In other words, a fixed $f$ can be homotopic to several points in the set of $S$-invariants functions. Of course what you’re saying is still correct, but basically the set of solutions of the problem is, to me, the subspace of $S$-invariants functions which are homotopic to $f$, I guess modulo $S$-equivariant homotopies. So in this sense the solution is not in general unique, which again is to be expected since it’s essentially equivalent to the pb of extending the morphism $\pi_1(X)\rightarrow$\pi_1(Z)$induced by$f$to a morphism$\pi_1(Y) \rightarrow \pi_1(Z)$along the inclusion induced by$p\$, and this problem doesn’t have a unique solution in general.

In case the covering is not regular, I also agree this should have something to do with homotopy fixed points for the $\pi_1(Y)$ action on the space fo maps $X\rightarrow Z$….

• CommentRowNumber4.
• CommentTimeOct 16th 2021
• (edited Oct 16th 2021)

[Sorry got logged out while I was typing… now i can’t edit, feel free to erase the previous message]

Hi Urs,

thanks a lot for your comment. The thing is it’s precisely not a principal bundle for some discrete group, because in my case it’s not a regular covering. In other words, $\pi_1(X)$ is not normal in $\pi_1(Y)$ (and indeed the particular situation I have a hard time NPOV-ing is the situation where its normal closure is the whole of $\pi_1(Y)$). That being said, even in the situation you describe there’s a small subtlety (I’m cutting hairs but it got me confused for a while). Namely it’s not quite true that $f$ factors uniquely, in the sense that I can have two maps $s,t:Y\rightarrow Z$ which are not homotopic, but such that their precomposition with $p$ are both homotopic to $f$. In other words, a fixed $f$ can be homotopic to several points in the space of $S$-invariants functions, which are however not themselves $S$-equivariantly homotopic.

Of course what you’re saying is still correct, but basically the set of solutions of the problem is, to me, the subspace of $S$-invariants functions which are homotopic to $f$, I guess modulo $S$-equivariant homotopies. So in this sense the solution is not in general unique, which again is to be expected since it’s essentially equivalent to the pb of extending the morphism $\pi_1(X)\rightarrow \pi_1(Z)$ induced by $f$ to a morphism $\pi_1(Y) \rightarrow \pi_1(Z)$ along the inclusion induced by $p$, and this problem doesn’t have a unique solution in general. i guess all i’m saying is that although the map $Map(Y,Z)=Map_S(X,Z)\rightarrow Map(X,Z)$ is injective exactly for the reasons you explained, the induced map between their $\pi_0$ is not.

In case the covering is not regular, I also agree this should have something to do with homotopy fixed points for the $\pi_1(Y)$ action on the space fo maps $X\rightarrow Z$….

1. I think I’ve isolated the claim doing what I want, does anyone know a reference for this ? Let $\rho:G\rightarrow H$ be a morphism of groups, and assume that $H$ is normally generated by $\im \rho$. Then the induced map $*/G \rightarrow */H$ is in fact an epimorphism in the (2,1)-category of groupoids. In particular I believe a covering map $X \rightarrow Y$ such that $\pi_1(X)$ normally generates $\pi_1(Y)$ should be an epimorphism in the $(\infty,1)$ category of spaces..

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeOct 18th 2021

I see. Sorry, I don’t know.

I never thought much about $\infty$-epimorphisms. But you are now essentially asking for the simplest non-trivial examples of these, and one would hope that somebody has thought about that.

Looking at the couple of references we have in that entry, Raptis 2017 does connect $\infty$-epimorphisms to conditions on normal subgroups, e.g in Lemma 3.4.

(This just by scanning for keywords, I don’t know if this is relevant for your question.)

2. Indeed, I've seen that reference and will look at it, thanks ! That being said my previous claim does not seem to be true, which is confusing...